a) Phân thức \(A\) xác định khi: \(x^2-25\ne0\Leftrightarrow x\ne\pm5\)

b) \(A=\dfrac{x^2-10x+25}{x^2-25}\)

\(=\dfrac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x-5}{x+5}\)

c) Với \(x=3\Rightarrow tmdk\)

Thay \(x=3\) vào \(A\), ta được

\(A=\dfrac{3-5}{3+5}=-\dfrac{1}{4}\)

d) Với \(x\ne\pm5\): \(A=\dfrac{1}{2}\Leftrightarrow\dfrac{x-5}{x+5}=\dfrac{1}{2}\)

\(\Leftrightarrow2x-10=x+5\)

\(\Leftrightarrow2x-x=5+10\)

\(\Leftrightarrow x=15\left(tmdk\right)\)

\(\text{#}Toru\)

\(a,ĐKXĐ:x^2-25\ne0\Leftrightarrow\left(x-5\right)\left(x+5\right)\ne0\Leftrightarrow x\ne\pm5\\ b,A=\dfrac{x^2-10x+25}{x^2-25}=\dfrac{x^2-2.x.5+5^2}{\left(x-5\right)\left(x+5\right)}=\dfrac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{x+5}\\ c,x=3\Rightarrow A=\dfrac{x-5}{x+5}=\dfrac{3-5}{3+5}=\dfrac{-2}{8}=-\dfrac{1}{4}\\ Vậy:x=3\Rightarrow A=-\dfrac{1}{4}\\ d,A=\dfrac{1}{2}\Leftrightarrow\dfrac{x-5}{x+5}=\dfrac{1}{2}\\ \Leftrightarrow2\left(x-5\right)=x+5\\ \Leftrightarrow2x-10=x+5\\ \Leftrightarrow2x-x=5+10\\ \Leftrightarrow x=15\\ Vậy:A=\dfrac{1}{2}\Leftrightarrow x=15\)

\(D=\dfrac{2ax-2x-3y+3ay}{4ax+6x+9y+6ay}=\dfrac{2x\left(a-1\right)+3y\left(a-1\right)}{2x\left(2a+3\right)+3y\left(2a+3\right)}\\ =\dfrac{\left(2x+3y\right)\left(a-1\right)}{\left(2x+3y\right)\left(2a+3\right)}=\dfrac{a-1}{2a+3}\)

Vậy GT biểu thức D không phụ thuộc vào x và y

\(C=\dfrac{\left(x+a\right)^2-x^2}{2x+a}=\dfrac{\left(x+a-x\right)\left(x+a+x\right)}{2x+a}\\ =\dfrac{a.\left(2x+a\right)}{2x+a}=a\)

Vậy GT biểu thức C không phụ thuộc vào x và y

\(a,\dfrac{x^5-2x^4+x^3}{x^4-2x^3+x^2}=\dfrac{x^3\left(x^2-2x+1\right)}{x^2\left(x^2-2x+1\right)}=\dfrac{x.x^2\left(x-1\right)^2}{x^2\left(x-1\right)^2}=x\left(đpcm\right)\\ b,\dfrac{4x^3-8x^2-x+2}{2x+1}=\dfrac{4x^2\left(x-2\right)-\left(x-2\right)}{2x+1}\\ =\dfrac{\left(4x^2-1\right)\left(x-2\right)}{2x+1}=\dfrac{\left(2x+1\right)\left(2x-1\right)\left(x-2\right)}{2x+1}=\left(2x-1\right)\left(x-2\right)\left(đpcm\right)\)

\(a,VT=\dfrac{x^5-2x^4+x^3}{x^4-2x^3+x^2}=\dfrac{x\left(x^4-2x^3+x^2\right)}{x^4-2x^3+x^2}=x=VP\left(x\ne0;x\ne1\right)\)

\(b,VT=\dfrac{4x^3-8x^2-x+2}{2x+1}\left(x\ne-\dfrac{1}{2}\right)\)

\(=\dfrac{4x^2\left(x-2\right)-\left(x-2\right)}{2x+1}\)

\(=\dfrac{\left(4x^2-1\right)\left(x-2\right)}{2x+1}\)

\(=\dfrac{\left(2x-1\right)\left(2x+1\right)\left(x-2\right)}{2x+1}\)

\(=\left(2x-1\right)\left(x-2\right)=VP\)

\(\text{#}Toru\)

a) \(\dfrac{3}{4ab^2}=\dfrac{3\cdot5a}{4ab^2\cdot5a}=\dfrac{15a}{20a^2b^2}\)

\(\dfrac{4}{5a^2b}=\dfrac{4\cdot4b}{5a^2b\cdot4b}=\dfrac{16b}{20a^2b^2}\)

b) \(\dfrac{x}{2x+6}=\dfrac{x}{2\left(x+3\right)}=\dfrac{x\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}\)

\(\dfrac{4}{x^2-9}=\dfrac{4}{\left(x+3\right)\left(x-3\right)}=\dfrac{8}{2\left(x+3\right)\left(x-3\right)}\)

c) \(\dfrac{2x}{x^3-1}=\dfrac{2x}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\dfrac{x+1}{x^2+x+1}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

d) \(\dfrac{x}{x^2+2x-3}=\dfrac{x}{\left(x-1\right)\left(x+3\right)}=\dfrac{x\left(x-1\right)}{\left(x-1\right)^2\left(x+3\right)}\)

\(\dfrac{3x}{x^2-2x+1}=\dfrac{3x}{\left(x-1\right)^2}=\dfrac{3x\left(x+3\right)}{\left(x-1\right)^2\left(x+3\right)}\)

\(\dfrac{-4}{x+3}=\dfrac{-4\left(x-1\right)^2}{\left(x-1\right)^2\left(x+3\right)}\)

a) \(M=\dfrac{15x^2y^3}{30xy^3z^2}=\dfrac{15xy^3\cdot x}{15xy^3\cdot2z^2}=\dfrac{x}{2z^2}\left(x;y;z\ne0\right)\)

b) \(N=\dfrac{x^4-4x^2}{x^2-4}=\dfrac{x^2\left(x^2-4\right)}{x^2-4}=x^2\left(x\ne\pm2\right)\)

c) \(P=\dfrac{x^3-3x^2+3x-1}{x^2-2x+1}=\dfrac{x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3}{x^2-2\cdot x\cdot1+1^2}=\dfrac{\left(x-1\right)^3}{\left(x-1\right)^2}=x-1\left(x\ne1\right)\)

d) \(A=\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x^2-4\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x+2\right)\left(x-2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\left(x\ne0;x\ne\pm2;x\ne-1\right)\)e) \(B=\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\left(x\ne-y\right)\) 

 a) Sửa đề: Chứng minh PF // DQ

Do CDEF là hình bình hành (gt)

⇒ CD // EF

⇒ DP // FQ

Tứ giác DPFQ có:

DP // FQ (cmt)

DP = FQ (gt)

⇒ DPFQ là hình bình hành

⇒ PF // DQ

b) Do CDEF là hình bình hành (gt)

O là trung điểm của CE (gt)

⇒ O là trung điểm của DF

Mà DPFQ là hình bình hành

⇒ O là trung điểm của PQ

Vườn ông minh là hình vuông 

Cạnh hình vuông là: 20 : 4 = 5 (m)

Chiều rộng hình chữ nhật sau khi tăng là: 5 + y (m)

Chiều dài hình chữ nhật sau khi tăng là: 8\(x\)  (m)

Diện tích hình chữ nhật sau khi tăng là: 8\(x\).(5 + y) (m²)

Ta có:

\(M=x^2-2x\left(y+1\right)+3y^2+2025\)

\(M=x^2-2\cdot x\cdot\left(y+1\right)+\left(y+1\right)^2+3y^2+2025-\left(y+1\right)^2\) 

\(M=\left[x-\left(y+1\right)\right]^2+3y^2+2025-y^2-2y-1\)

\(M=\left(x-y-1\right)^2+2y^2-2y+2024\)

\(M=\left(x-y-1\right)^2+2\left(y-\dfrac{1}{2}\right)^2+\dfrac{4047}{2}\)

Mà: \(\left\{{}\begin{matrix}\left(x-y-1\right)^2\ge0\\2\left(y-\dfrac{1}{2}\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow M=\left(x-y-1\right)^2+2\left(y-\dfrac{1}{2}\right)^2+\dfrac{4047}{2}\ge\dfrac{4047}{2}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x-y-1=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}+1\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\) 

Vậy GTNN của M là .... 

Yêu cầu đề là gì vậy bạn?

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